/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *    http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

/*
 * Based on TimSort.java from the Android Open Source Project
 *
 *  Copyright (C) 2008 The Android Open Source Project
 *
 *  Licensed under the Apache License, Version 2.0 (the "License");
 *  you may not use this file except in compliance with the License.
 *  You may obtain a copy of the License at
 *
 *       http://www.apache.org/licenses/LICENSE-2.0
 *
 *  Unless required by applicable law or agreed to in writing, software
 *  distributed under the License is distributed on an "AS IS" BASIS,
 *  WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 *  See the License for the specific language governing permissions and
 *  limitations under the License.
 */

package org.apache.spark.util.collection;

import java.util.Comparator;

/**
 * A port of the Android TimSort class, which utilizes a "stable, adaptive, iterative mergesort."
 * See the method comment on sort() for more details.
 *
 * This has been kept in Java with the original style in order to match very closely with the
 * Android source code, and thus be easy to verify correctness. The class is package private. We put
 * a simple Scala wrapper {@link org.apache.spark.util.collection.Sorter}, which is available to
 * package org.apache.spark.
 *
 * The purpose of the port is to generalize the interface to the sort to accept input data formats
 * besides simple arrays where every element is sorted individually. For instance, the AppendOnlyMap
 * uses this to sort an Array with alternating elements of the form [key, value, key, value].
 * This generalization comes with minimal overhead -- see SortDataFormat for more information.
 *
 * We allow key reuse to prevent creating many key objects -- see SortDataFormat.
 *
 * @see org.apache.spark.util.collection.SortDataFormat
 * @see org.apache.spark.util.collection.Sorter
 */
public final class TimSort<K, Buffer> {

    /**
     * This is the minimum sized sequence that will be merged.  Shorter
     * sequences will be lengthened by calling binarySort.  If the entire
     * array is less than this length, no merges will be performed.
     *
     * This constant should be a power of two.  It was 64 in Tim Peter's C
     * implementation, but 32 was empirically determined to work better in
     * this implementation.  In the unlikely event that you set this constant
     * to be a number that's not a power of two, you'll need to change the
     * minRunLength computation.
     *
     * If you decrease this constant, you must change the stackLen
     * computation in the TimSort constructor, or you risk an
     * ArrayOutOfBounds exception.  See listsort.txt for a discussion
     * of the minimum stack length required as a function of the length
     * of the array being sorted and the minimum merge sequence length.
     */
    private static final int MIN_MERGE = 32;

    private final SortDataFormat<K, Buffer> s;

    public TimSort(SortDataFormat<K, Buffer> sortDataFormat) {
        this.s = sortDataFormat;
    }

    /**
     * A stable, adaptive, iterative mergesort that requires far fewer than
     * n lg(n) comparisons when running on partially sorted arrays, while
     * offering performance comparable to a traditional mergesort when run
     * on random arrays.  Like all proper mergesorts, this sort is stable and
     * runs O(n log n) time (worst case).  In the worst case, this sort requires
     * temporary storage space for n/2 object references; in the best case,
     * it requires only a small constant amount of space.
     *
     * This implementation was adapted from Tim Peters's list sort for
     * Python, which is described in detail here:
     *
     * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
     *
     * Tim's C code may be found here:
     *
     * http://svn.python.org/projects/python/trunk/Objects/listobject.c
     *
     * The underlying techniques are described in this paper (and may have
     * even earlier origins):
     *
     * "Optimistic Sorting and Information Theoretic Complexity"
     * Peter McIlroy
     * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
     * pp 467-474, Austin, Texas, 25-27 January 1993.
     *
     * While the API to this class consists solely of static methods, it is
     * (privately) instantiable; a TimSort instance holds the state of an ongoing
     * sort, assuming the input array is large enough to warrant the full-blown
     * TimSort. Small arrays are sorted in place, using a binary insertion sort.
     *
     * @author Josh Bloch
     */
    public void sort(Buffer a, int lo, int hi, Comparator<? super K> c) {
        assert c != null;

        int nRemaining = hi - lo;
        if (nRemaining < 2)
            return;  // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        if (nRemaining < MIN_MERGE) {
            int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
            binarySort(a, lo, hi, lo + initRunLen, c);
            return;
        }

        /**
         * March over the array once, left to right, finding natural runs,
         * extending short natural runs to minRun elements, and merging runs
         * to maintain stack invariant.
         */
        SortState sortState = new SortState(a, c, hi - lo);
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            int runLen = countRunAndMakeAscending(a, lo, hi, c);

            // If run is short, extend to min(minRun, nRemaining)
            if (runLen < minRun) {
                int force = nRemaining <= minRun ? nRemaining : minRun;
                binarySort(a, lo, lo + force, lo + runLen, c);
                runLen = force;
            }

            // Push run onto pending-run stack, and maybe merge
            sortState.pushRun(lo, runLen);
            sortState.mergeCollapse();

            // Advance to find next run
            lo += runLen;
            nRemaining -= runLen;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        assert lo == hi;
        sortState.mergeForceCollapse();
        assert sortState.stackSize == 1;
    }

    /**
     * Sorts the specified portion of the specified array using a binary
     * insertion sort.  This is the best method for sorting small numbers
     * of elements.  It requires O(n log n) compares, but O(n^2) data
     * movement (worst case).
     *
     * If the initial part of the specified range is already sorted,
     * this method can take advantage of it: the method assumes that the
     * elements from index {@code lo}, inclusive, to {@code start},
     * exclusive are already sorted.
     *
     * @param a     the array in which a range is to be sorted
     * @param lo    the index of the first element in the range to be sorted
     * @param hi    the index after the last element in the range to be sorted
     * @param start the index of the first element in the range that is
     *              not already known to be sorted ({@code lo <= start <= hi})
     * @param c     comparator to used for the sort
     */
    @SuppressWarnings("fallthrough")
    private void binarySort(Buffer a, int lo, int hi, int start, Comparator<? super K> c) {
        assert lo <= start && start <= hi;
        if (start == lo)
            start++;

        K key0 = s.newKey();
        K key1 = s.newKey();

        Buffer pivotStore = s.allocate(1);
        for (; start < hi; start++) {
            s.copyElement(a, start, pivotStore, 0);
            K pivot = s.getKey(pivotStore, 0, key0);

            // Set left (and right) to the index where a[start] (pivot) belongs
            int left = lo;
            int right = start;
            assert left <= right;
      /*
       * Invariants:
       *   pivot >= all in [lo, left).
       *   pivot <  all in [right, start).
       */
            while (left < right) {
                int mid = (left + right) >>> 1;
                if (c.compare(pivot, s.getKey(a, mid, key1)) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            assert left == right;

      /*
       * The invariants still hold: pivot >= all in [lo, left) and
       * pivot < all in [left, start), so pivot belongs at left.  Note
       * that if there are elements equal to pivot, left points to the
       * first slot after them -- that's why this sort is stable.
       * Slide elements over to make room for pivot.
       */
            int n = start - left;  // The number of elements to move
            // Switch is just an optimization for arraycopy in default case
            switch (n) {
                case 2:
                    s.copyElement(a, left + 1, a, left + 2);
                case 1:
                    s.copyElement(a, left, a, left + 1);
                    break;
                default:
                    s.copyRange(a, left, a, left + 1, n);
            }
            s.copyElement(pivotStore, 0, a, left);
        }
    }

    /**
     * Returns the length of the run beginning at the specified position in
     * the specified array and reverses the run if it is descending (ensuring
     * that the run will always be ascending when the method returns).
     *
     * A run is the longest ascending sequence with:
     *
     * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
     *
     * or the longest descending sequence with:
     *
     * a[lo] >  a[lo + 1] >  a[lo + 2] >  ...
     *
     * For its intended use in a stable mergesort, the strictness of the
     * definition of "descending" is needed so that the call can safely
     * reverse a descending sequence without violating stability.
     *
     * @param a  the array in which a run is to be counted and possibly reversed
     * @param lo index of the first element in the run
     * @param hi index after the last element that may be contained in the run.
     *           It is required that {@code lo < hi}.
     * @param c  the comparator to used for the sort
     * @return the length of the run beginning at the specified position in
     * the specified array
     */
    private int countRunAndMakeAscending(Buffer a, int lo, int hi, Comparator<? super K> c) {
        assert lo < hi;
        int runHi = lo + 1;
        if (runHi == hi)
            return 1;

        K key0 = s.newKey();
        K key1 = s.newKey();

        // Find end of run, and reverse range if descending
        if (c.compare(s.getKey(a, runHi++, key0), s.getKey(a, lo, key1)) < 0) { // Descending
            while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) < 0)
                runHi++;
            reverseRange(a, lo, runHi);
        } else {                              // Ascending
            while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) >= 0)
                runHi++;
        }

        return runHi - lo;
    }

    /**
     * Reverse the specified range of the specified array.
     *
     * @param a  the array in which a range is to be reversed
     * @param lo the index of the first element in the range to be reversed
     * @param hi the index after the last element in the range to be reversed
     */
    private void reverseRange(Buffer a, int lo, int hi) {
        hi--;
        while (lo < hi) {
            s.swap(a, lo, hi);
            lo++;
            hi--;
        }
    }

    /**
     * Returns the minimum acceptable run length for an array of the specified
     * length. Natural runs shorter than this will be extended with
     * {@link #binarySort}.
     *
     * Roughly speaking, the computation is:
     *
     * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
     * Else if n is an exact power of 2, return MIN_MERGE/2.
     * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
     * is close to, but strictly less than, an exact power of 2.
     *
     * For the rationale, see listsort.txt.
     *
     * @param n the length of the array to be sorted
     * @return the length of the minimum run to be merged
     */
    private int minRunLength(int n) {
        assert n >= 0;
        int r = 0;      // Becomes 1 if any 1 bits are shifted off
        while (n >= MIN_MERGE) {
            r |= (n & 1);
            n >>= 1;
        }
        return n + r;
    }

    private class SortState {

        /**
         * The Buffer being sorted.
         */
        private final Buffer a;

        /**
         * Length of the sort Buffer.
         */
        private final int aLength;

        /**
         * The comparator for this sort.
         */
        private final Comparator<? super K> c;

        /**
         * When we get into galloping mode, we stay there until both runs win less
         * often than MIN_GALLOP consecutive times.
         */
        private static final int MIN_GALLOP = 7;

        /**
         * This controls when we get *into* galloping mode.  It is initialized
         * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
         * random data, and lower for highly structured data.
         */
        private int minGallop = MIN_GALLOP;

        /**
         * Maximum initial size of tmp array, which is used for merging.  The array
         * can grow to accommodate demand.
         *
         * Unlike Tim's original C version, we do not allocate this much storage
         * when sorting smaller arrays.  This change was required for performance.
         */
        private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

        /**
         * Temp storage for merges.
         */
        private Buffer tmp; // Actual runtime type will be Object[], regardless of T

        /**
         * Length of the temp storage.
         */
        private int tmpLength = 0;

        /**
         * A stack of pending runs yet to be merged.  Run i starts at
         * address base[i] and extends for len[i] elements.  It's always
         * true (so long as the indices are in bounds) that:
         *
         * runBase[i] + runLen[i] == runBase[i + 1]
         *
         * so we could cut the storage for this, but it's a minor amount,
         * and keeping all the info explicit simplifies the code.
         */
        private int stackSize = 0;  // Number of pending runs on stack
        private final int[] runBase;
        private final int[] runLen;

        /**
         * Creates a TimSort instance to maintain the state of an ongoing sort.
         *
         * @param a the array to be sorted
         * @param c the comparator to determine the order of the sort
         */
        private SortState(Buffer a, Comparator<? super K> c, int len) {
            this.aLength = len;
            this.a = a;
            this.c = c;

            // Allocate temp storage (which may be increased later if necessary)
            tmpLength = len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
            tmp = s.allocate(tmpLength);

      /*
       * Allocate runs-to-be-merged stack (which cannot be expanded).  The
       * stack length requirements are described in listsort.txt.  The C
       * version always uses the same stack length (85), but this was
       * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
       * 100 elements) in Java.  Therefore, we use smaller (but sufficiently
       * large) stack lengths for smaller arrays.  The "magic numbers" in the
       * computation below must be changed if MIN_MERGE is decreased.  See
       * the MIN_MERGE declaration above for more information.
       */
            int stackLen = (len < 120 ? 5 :
                    len < 1542 ? 10 :
                            len < 119151 ? 19 : 40);
            runBase = new int[stackLen];
            runLen = new int[stackLen];
        }

        /**
         * Pushes the specified run onto the pending-run stack.
         *
         * @param runBase index of the first element in the run
         * @param runLen  the number of elements in the run
         */
        private void pushRun(int runBase, int runLen) {
            this.runBase[stackSize] = runBase;
            this.runLen[stackSize] = runLen;
            stackSize++;
        }

        /**
         * Examines the stack of runs waiting to be merged and merges adjacent runs
         * until the stack invariants are reestablished:
         *
         * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
         * 2. runLen[i - 2] > runLen[i - 1]
         *
         * This method is called each time a new run is pushed onto the stack,
         * so the invariants are guaranteed to hold for i < stackSize upon
         * entry to the method.
         */
        private void mergeCollapse() {
            while (stackSize > 1) {
                int n = stackSize - 2;
                if ((n >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1])
                        || (n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1])) {
                    if (runLen[n - 1] < runLen[n + 1])
                        n--;
                } else if (runLen[n] > runLen[n + 1]) {
                    break; // Invariant is established
                }
                mergeAt(n);
            }
        }

        /**
         * Merges all runs on the stack until only one remains.  This method is
         * called once, to complete the sort.
         */
        private void mergeForceCollapse() {
            while (stackSize > 1) {
                int n = stackSize - 2;
                if (n > 0 && runLen[n - 1] < runLen[n + 1])
                    n--;
                mergeAt(n);
            }
        }

        /**
         * Merges the two runs at stack indices i and i+1.  Run i must be
         * the penultimate or antepenultimate run on the stack.  In other words,
         * i must be equal to stackSize-2 or stackSize-3.
         *
         * @param i stack index of the first of the two runs to merge
         */
        private void mergeAt(int i) {
            assert stackSize >= 2;
            assert i >= 0;
            assert i == stackSize - 2 || i == stackSize - 3;

            int base1 = runBase[i];
            int len1 = runLen[i];
            int base2 = runBase[i + 1];
            int len2 = runLen[i + 1];
            assert len1 > 0 && len2 > 0;
            assert base1 + len1 == base2;

      /*
       * Record the length of the combined runs; if i is the 3rd-last
       * run now, also slide over the last run (which isn't involved
       * in this merge).  The current run (i+1) goes away in any case.
       */
            runLen[i] = len1 + len2;
            if (i == stackSize - 3) {
                runBase[i + 1] = runBase[i + 2];
                runLen[i + 1] = runLen[i + 2];
            }
            stackSize--;

            K key0 = s.newKey();

      /*
       * Find where the first element of run2 goes in run1. Prior elements
       * in run1 can be ignored (because they're already in place).
       */
            int k = gallopRight(s.getKey(a, base2, key0), a, base1, len1, 0, c);
            assert k >= 0;
            base1 += k;
            len1 -= k;
            if (len1 == 0)
                return;

      /*
       * Find where the last element of run1 goes in run2. Subsequent elements
       * in run2 can be ignored (because they're already in place).
       */
            len2 = gallopLeft(s.getKey(a, base1 + len1 - 1, key0), a, base2, len2, len2 - 1, c);
            assert len2 >= 0;
            if (len2 == 0)
                return;

            // Merge remaining runs, using tmp array with min(len1, len2) elements
            if (len1 <= len2)
                mergeLo(base1, len1, base2, len2);
            else
                mergeHi(base1, len1, base2, len2);
        }

        /**
         * Locates the position at which to insert the specified key into the
         * specified sorted range; if the range contains an element equal to key,
         * returns the index of the leftmost equal element.
         *
         * @param key  the key whose insertion point to search for
         * @param a    the array in which to search
         * @param base the index of the first element in the range
         * @param len  the length of the range; must be > 0
         * @param hint the index at which to begin the search, 0 <= hint < n.
         *             The closer hint is to the result, the faster this method will run.
         * @param c    the comparator used to order the range, and to search
         * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
         * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
         * In other words, key belongs at index b + k; or in other words,
         * the first k elements of a should precede key, and the last n - k
         * should follow it.
         */
        private int gallopLeft(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
            assert len > 0 && hint >= 0 && hint < len;
            int lastOfs = 0;
            int ofs = 1;
            K key0 = s.newKey();

            if (c.compare(key, s.getKey(a, base + hint, key0)) > 0) {
                // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
                int maxOfs = len - hint;
                while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key0)) > 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0)   // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs)
                    ofs = maxOfs;

                // Make offsets relative to base
                lastOfs += hint;
                ofs += hint;
            } else { // key <= a[base + hint]
                // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
                final int maxOfs = hint + 1;
                while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key0)) <= 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0)   // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs)
                    ofs = maxOfs;

                // Make offsets relative to base
                int tmp = lastOfs;
                lastOfs = hint - ofs;
                ofs = hint - tmp;
            }
            assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

      /*
       * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
       * to the right of lastOfs but no farther right than ofs.  Do a binary
       * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
       */
            lastOfs++;
            while (lastOfs < ofs) {
                int m = lastOfs + ((ofs - lastOfs) >>> 1);

                if (c.compare(key, s.getKey(a, base + m, key0)) > 0)
                    lastOfs = m + 1;  // a[base + m] < key
                else
                    ofs = m;          // key <= a[base + m]
            }
            assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
            return ofs;
        }

        /**
         * Like gallopLeft, except that if the range contains an element equal to
         * key, gallopRight returns the index after the rightmost equal element.
         *
         * @param key  the key whose insertion point to search for
         * @param a    the array in which to search
         * @param base the index of the first element in the range
         * @param len  the length of the range; must be > 0
         * @param hint the index at which to begin the search, 0 <= hint < n.
         *             The closer hint is to the result, the faster this method will run.
         * @param c    the comparator used to order the range, and to search
         * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
         */
        private int gallopRight(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
            assert len > 0 && hint >= 0 && hint < len;

            int ofs = 1;
            int lastOfs = 0;
            K key1 = s.newKey();

            if (c.compare(key, s.getKey(a, base + hint, key1)) < 0) {
                // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
                int maxOfs = hint + 1;
                while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key1)) < 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0)   // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs)
                    ofs = maxOfs;

                // Make offsets relative to b
                int tmp = lastOfs;
                lastOfs = hint - ofs;
                ofs = hint - tmp;
            } else { // a[b + hint] <= key
                // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
                int maxOfs = len - hint;
                while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key1)) >= 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0)   // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs)
                    ofs = maxOfs;

                // Make offsets relative to b
                lastOfs += hint;
                ofs += hint;
            }
            assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

      /*
       * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
       * the right of lastOfs but no farther right than ofs.  Do a binary
       * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
       */
            lastOfs++;
            while (lastOfs < ofs) {
                int m = lastOfs + ((ofs - lastOfs) >>> 1);

                if (c.compare(key, s.getKey(a, base + m, key1)) < 0)
                    ofs = m;          // key < a[b + m]
                else
                    lastOfs = m + 1;  // a[b + m] <= key
            }
            assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
            return ofs;
        }

        /**
         * Merges two adjacent runs in place, in a stable fashion.  The first
         * element of the first run must be greater than the first element of the
         * second run (a[base1] > a[base2]), and the last element of the first run
         * (a[base1 + len1-1]) must be greater than all elements of the second run.
         *
         * For performance, this method should be called only when len1 <= len2;
         * its twin, mergeHi should be called if len1 >= len2.  (Either method
         * may be called if len1 == len2.)
         *
         * @param base1 index of first element in first run to be merged
         * @param len1  length of first run to be merged (must be > 0)
         * @param base2 index of first element in second run to be merged
         *              (must be aBase + aLen)
         * @param len2  length of second run to be merged (must be > 0)
         */
        private void mergeLo(int base1, int len1, int base2, int len2) {
            assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

            // Copy first run into temp array
            Buffer a = this.a; // For performance
            Buffer tmp = ensureCapacity(len1);
            s.copyRange(a, base1, tmp, 0, len1);

            int cursor1 = 0;       // Indexes into tmp array
            int cursor2 = base2;   // Indexes int a
            int dest = base1;      // Indexes int a

            // Move first element of second run and deal with degenerate cases
            s.copyElement(a, cursor2++, a, dest++);
            if (--len2 == 0) {
                s.copyRange(tmp, cursor1, a, dest, len1);
                return;
            }
            if (len1 == 1) {
                s.copyRange(a, cursor2, a, dest, len2);
                s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run 1 to end of merge
                return;
            }

            K key0 = s.newKey();
            K key1 = s.newKey();

            Comparator<? super K> c = this.c;  // Use local variable for performance
            int minGallop = this.minGallop;    //  "    "       "     "      "
            outer:
            while (true) {
                int count1 = 0; // Number of times in a row that first run won
                int count2 = 0; // Number of times in a row that second run won

        /*
         * Do the straightforward thing until (if ever) one run starts
         * winning consistently.
         */
                do {
                    assert len1 > 1 && len2 > 0;
                    if (c.compare(s.getKey(a, cursor2, key0), s.getKey(tmp, cursor1, key1)) < 0) {
                        s.copyElement(a, cursor2++, a, dest++);
                        count2++;
                        count1 = 0;
                        if (--len2 == 0)
                            break outer;
                    } else {
                        s.copyElement(tmp, cursor1++, a, dest++);
                        count1++;
                        count2 = 0;
                        if (--len1 == 1)
                            break outer;
                    }
                } while ((count1 | count2) < minGallop);

        /*
         * One run is winning so consistently that galloping may be a
         * huge win. So try that, and continue galloping until (if ever)
         * neither run appears to be winning consistently anymore.
         */
                do {
                    assert len1 > 1 && len2 > 0;
                    count1 = gallopRight(s.getKey(a, cursor2, key0), tmp, cursor1, len1, 0, c);
                    if (count1 != 0) {
                        s.copyRange(tmp, cursor1, a, dest, count1);
                        dest += count1;
                        cursor1 += count1;
                        len1 -= count1;
                        if (len1 <= 1) // len1 == 1 || len1 == 0
                            break outer;
                    }
                    s.copyElement(a, cursor2++, a, dest++);
                    if (--len2 == 0)
                        break outer;

                    count2 = gallopLeft(s.getKey(tmp, cursor1, key0), a, cursor2, len2, 0, c);
                    if (count2 != 0) {
                        s.copyRange(a, cursor2, a, dest, count2);
                        dest += count2;
                        cursor2 += count2;
                        len2 -= count2;
                        if (len2 == 0)
                            break outer;
                    }
                    s.copyElement(tmp, cursor1++, a, dest++);
                    if (--len1 == 1)
                        break outer;
                    minGallop--;
                } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
                if (minGallop < 0)
                    minGallop = 0;
                minGallop += 2;  // Penalize for leaving gallop mode
            }  // End of "outer" loop
            this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

            if (len1 == 1) {
                assert len2 > 0;
                s.copyRange(a, cursor2, a, dest, len2);
                s.copyElement(tmp, cursor1, a, dest + len2); //  Last elt of run 1 to end of merge
            } else if (len1 == 0) {
                throw new IllegalArgumentException(
                        "Comparison method violates its general contract!");
            } else {
                assert len2 == 0;
                assert len1 > 1;
                s.copyRange(tmp, cursor1, a, dest, len1);
            }
        }

        /**
         * Like mergeLo, except that this method should be called only if
         * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
         * may be called if len1 == len2.)
         *
         * @param base1 index of first element in first run to be merged
         * @param len1  length of first run to be merged (must be > 0)
         * @param base2 index of first element in second run to be merged
         *              (must be aBase + aLen)
         * @param len2  length of second run to be merged (must be > 0)
         */
        private void mergeHi(int base1, int len1, int base2, int len2) {
            assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

            // Copy second run into temp array
            Buffer a = this.a; // For performance
            Buffer tmp = ensureCapacity(len2);
            s.copyRange(a, base2, tmp, 0, len2);

            int cursor1 = base1 + len1 - 1;  // Indexes into a
            int cursor2 = len2 - 1;          // Indexes into tmp array
            int dest = base2 + len2 - 1;     // Indexes into a

            K key0 = s.newKey();
            K key1 = s.newKey();

            // Move last element of first run and deal with degenerate cases
            s.copyElement(a, cursor1--, a, dest--);
            if (--len1 == 0) {
                s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
                return;
            }
            if (len2 == 1) {
                dest -= len1;
                cursor1 -= len1;
                s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
                s.copyElement(tmp, cursor2, a, dest);
                return;
            }

            Comparator<? super K> c = this.c;  // Use local variable for performance
            int minGallop = this.minGallop;    //  "    "       "     "      "
            outer:
            while (true) {
                int count1 = 0; // Number of times in a row that first run won
                int count2 = 0; // Number of times in a row that second run won

        /*
         * Do the straightforward thing until (if ever) one run
         * appears to win consistently.
         */
                do {
                    assert len1 > 0 && len2 > 1;
                    if (c.compare(s.getKey(tmp, cursor2, key0), s.getKey(a, cursor1, key1)) < 0) {
                        s.copyElement(a, cursor1--, a, dest--);
                        count1++;
                        count2 = 0;
                        if (--len1 == 0)
                            break outer;
                    } else {
                        s.copyElement(tmp, cursor2--, a, dest--);
                        count2++;
                        count1 = 0;
                        if (--len2 == 1)
                            break outer;
                    }
                } while ((count1 | count2) < minGallop);

        /*
         * One run is winning so consistently that galloping may be a
         * huge win. So try that, and continue galloping until (if ever)
         * neither run appears to be winning consistently anymore.
         */
                do {
                    assert len1 > 0 && len2 > 1;
                    count1 = len1 - gallopRight(s.getKey(tmp, cursor2, key0), a, base1, len1, len1 - 1, c);
                    if (count1 != 0) {
                        dest -= count1;
                        cursor1 -= count1;
                        len1 -= count1;
                        s.copyRange(a, cursor1 + 1, a, dest + 1, count1);
                        if (len1 == 0)
                            break outer;
                    }
                    s.copyElement(tmp, cursor2--, a, dest--);
                    if (--len2 == 1)
                        break outer;

                    count2 = len2 - gallopLeft(s.getKey(a, cursor1, key0), tmp, 0, len2, len2 - 1, c);
                    if (count2 != 0) {
                        dest -= count2;
                        cursor2 -= count2;
                        len2 -= count2;
                        s.copyRange(tmp, cursor2 + 1, a, dest + 1, count2);
                        if (len2 <= 1)  // len2 == 1 || len2 == 0
                            break outer;
                    }
                    s.copyElement(a, cursor1--, a, dest--);
                    if (--len1 == 0)
                        break outer;
                    minGallop--;
                } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
                if (minGallop < 0)
                    minGallop = 0;
                minGallop += 2;  // Penalize for leaving gallop mode
            }  // End of "outer" loop
            this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

            if (len2 == 1) {
                assert len1 > 0;
                dest -= len1;
                cursor1 -= len1;
                s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
                s.copyElement(tmp, cursor2, a, dest); // Move first elt of run2 to front of merge
            } else if (len2 == 0) {
                throw new IllegalArgumentException(
                        "Comparison method violates its general contract!");
            } else {
                assert len1 == 0;
                assert len2 > 0;
                s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
            }
        }

        /**
         * Ensures that the external array tmp has at least the specified
         * number of elements, increasing its size if necessary.  The size
         * increases exponentially to ensure amortized linear time complexity.
         *
         * @param minCapacity the minimum required capacity of the tmp array
         * @return tmp, whether or not it grew
         */
        private Buffer ensureCapacity(int minCapacity) {
            if (tmpLength < minCapacity) {
                // Compute smallest power of 2 > minCapacity
                int newSize = minCapacity;
                newSize |= newSize >> 1;
                newSize |= newSize >> 2;
                newSize |= newSize >> 4;
                newSize |= newSize >> 8;
                newSize |= newSize >> 16;
                newSize++;

                if (newSize < 0) // Not bloody likely!
                    newSize = minCapacity;
                else
                    newSize = Math.min(newSize, aLength >>> 1);

                tmp = s.allocate(newSize);
                tmpLength = newSize;
            }
            return tmp;
        }
    }
}
